All categories

3 years ago

Complete the coordinate proof of the theorem.

Mathematics

2 answers:

Flauer [41]3 years ago

8 0

Answer:

Step-by-step explanation:

THESE ARE THE CORRECT ANSWERS

mojhsa [17]3 years ago

3 0

C(a,b), because the x-coordinate( first coordinate) is a (seeing as it is situated directly above point B, which also has an x-coordinate of a) and the y-coordinate ( second coordinate) is b (seeing as it is situated on the same horizontal level as point D, which also has a y-coordinate of b)

the length of AC can be calculated with the theorem of Pythagoras:

length AB = a - 0 = a
length BC = b - 0 = b

seeing as the length of AC is the longest, it can be calculated by the following formula:

It is called "Pythagoras' Theorem" and can be written in one short equation:

a^2 + b^2 = c^2 (^ means to the power of by the way)

in this case, A and B are lengths AB and BC, so lenght AC can be calculated as the following:

a^2 + b^2 = (length AC)^2
length AC = √(a^2 + b^2)

Extra information: Seeing as the shape of the drawn lines is a rectangle, lines AC and BD have to be the same length, so BD is also √(a^2 + b^2). But that is also stated in the assignment!

You might be interested in

How do I do this? it doesn't make sense.

yuradex [85]

I think you subtitute de 34 by x not sure

4 0

3 years ago

Round to nearest hundredth<br>0.099

Degger [83]

Answer: 0.10

Step-by-step explanation:

Find the number in the hundredth place

and look one place to the right for the rounding digit

. Round up if this number is greater than or equal to

and round down if it is less than

0.10

8 0

2 years ago

Kathy is fencing in her yard that has an area of 800 square feet. Her yard is in the shape of a rectangle where the length is 60

anygoal [31]

Answer:

The dimensions of the yard are W=20ft and L=40ft.

Step-by-step explanation:

Let be:

W: width of the yard.

L:length.

Now, we can write the equation of that relates length and width:

$L=5W-60$ (Equation #1)

The area of the yard can be expressed as (using equation #1 into #2):

$Area=W*L=W*(5W-60)$ (Equation #2)

Since the Area of the yard is $800 ft^2$ , then equation #2 turns into:

$800=W*(5W-60)$

Now, we rearrange this equation:

$800=W*(5W-60)//800=5W^2-60W//5W^2-60W-800=0$

We can divide the equation by 5 :

$W^2-12W-160=0$

We need to find the solution for this quadratic. Let's find the factors of 160 that multiplied yields -160 and added yields -12. Let's choose -20 and 8, since $(-20)*8=160$ and $-20+8=-12$ . The equation factorised looks like this:

$(W-20)(W+8)=0$

Therefore the possible solutions are W=20 and W=-8. We discard W=-8 since width must be a positive number. To find the length, we substitute the value of W in equation #1:

$5*20-60=40$